A projectile shot at an angle of $45^{o}$ above the horizontal strikes a building 30 m away at a point 15 m above the point of projection. Find : (a) the speed of projection. (b) the magnitude and direction of velocity of projectile when it strikes the building

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Solution

$y = x t a n θ_{o} - \frac{1}{2} \frac{8}{{u_{o}}^{2} c o s^{2} θ_{o}} x^{2}_____(1)$

$v^{2} = u^{2} - 2 g h$

Vertical component of velocity

${V_{x}}^{2} = [14 \sqrt{3} \times s i n 45]^{2} - 2 \times 9.8 \times 15$

$= 0$

We have only horizontal component

$V_{y} = 14 \sqrt{3} c o s 45$

$= 17.15 m s^{- 1}$ (IMAGE)

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A projectile shot at an angle of 45o above the horizontal strikes a building 30 m away at a point 15 m above the point of projection. Find : (a) the speed of projection. (b) the magnitude and direction of velocity of projectile when it strikes the building

y=xtanθo−128uo2cos2θox2_____(1) v2=u2–2gh Vertical component of velocity Vx2=[14√3× sin45]2–2×9.8×15 =0 We have only horizontal component Vy=14√3 cos45 =17.15ms−1 (IMAGE)

A projectile shot at an angle of $45^{o}$ above the horizontal strikes a building 30 m away at a point 15 m above the point of projection. Find : (a) the speed of projection. (b) the magnitude and direction of velocity of projectile when it strikes the building